Linear Algebra/Strings/Solutions
Solutions
edit- This exercise is recommended for all readers.
- Problem 1
What is the index of nilpotency of the left-shift operator, here acting on the space of triples of reals?
- Answer
Three. It is at least three because . It is at most three because .
- This exercise is recommended for all readers.
- Problem 2
For each string basis state the index of nilpotency and give the dimension of the rangespace and nullspace of each iteration of the nilpotent map.
Also give the canonical form of the matrix.
- Answer
- The domain has dimension four.
The map's action is that any vector in the space
is sent to
.
The first application of the map
sends two basis vectors and
to zero,
and therefore the nullspace has dimension two and the rangespace
has dimension two.
With a second application, all four basis vectors are sent to
zero and so the nullspace of the second power has dimension four
while the rangespace of the second power has dimension zero.
Thus the index of nilpotency is two.
This is the canonical form.
- The dimension of the domain of this map is six.
For the first power the dimension of the nullspace is four
and the dimension of the rangespace is two.
For the second power the dimension of the nullspace is five
and the dimension of the rangespace is one.
Then the third iteration results in a nullspace of dimension
six and a rangespace of dimension zero.
The index of nilpotency is three, and this is the
canonical form.
- The dimension of the domain is three, and the index of
nilpotency is three.
The first power's null space has dimension one and its range space
has dimension two.
The second power's null space has dimension two and its range space
has dimension one.
Finally, the third power's null space has dimension three
and its range space
has dimension zero.
Here is the canonical form matrix.
- Problem 3
Decide which of these matrices are nilpotent.
- Answer
By Lemma 1.3 the nullity has grown as large as possible by the -th iteration where is the dimension of the domain. Thus, for the matrices, we need only check whether the square is the zero matrix. For the matrices, we need only check the cube.
- Yes, this matrix is nilpotent because its square is the zero matrix.
- No, the square is not the zero matrix.
- Yes, the cube is the zero matrix. In fact, the square is zero.
- No, the third power is not the zero matrix.
- Yes, the cube of this matrix is the zero matrix.
Another way to see that the second and fourth matrices are not nilpotent is to note that they are nonsingular.
- This exercise is recommended for all readers.
- Problem 4
Find the canonical form of this matrix.
- Answer
The table os calculations
gives these requirements of the string basis: three basis vectors are sent directly to zero, one more basis vector is sent to zero by a second application, and the final basis vector is sent to zero by a third application. Thus, the string basis has this form.
From that the canonical form is immediate.
- This exercise is recommended for all readers.
- Problem 5
Consider the matrix from Example 2.16.
- Use the action of the map on the string basis to give the canonical form.
- Find the change of basis matrices that bring the matrix to canonical form.
- Use the answer in the prior item to check the answer in the first item.
- Answer
- The canonical form has a block and a
block
- Assume that is the representation of the underlying
map with respect to the standard basis.
Let be the basis to which we will change.
By the similarity diagram
we have that the canonical form matrix is where
- The calculation to check this is routine.
- This exercise is recommended for all readers.
- Problem 6
Each of these matrices is nilpotent.
Put each in canonical form.
- Answer
- The calculation
shows that any map represented by the matrix must act on the string basis in this way
because the nullspace after one application has dimension one and exactly one basis vector, , is sent to zero. Therefore, this representation with respect to is the canonical form.
- The calculation here is similar to the prior one.
The table shows that the string basis is of the form
because the nullspace after one application of the map has dimension two— and are both sent to zero— and one more iteration results in the additional vector being brought to zero.
- The calculation
shows that any map represented by this basis must act on a string basis in this way.
Therefore, this is the canonical form.
- Problem 7
Describe the effect of left or right multiplication by a matrix that is in the canonical form for nilpotent matrices.
- Answer
A couple of examples
suggest that left multiplication by a block of subdiagonal ones shifts the rows of a matrix downward. Distinct blocks
act to shift down distinct parts of the matrix.
Right multiplication does an analagous thing to columns. See Problem 1.
- Problem 8
Is nilpotence invariant under similarity? That is, must a matrix similar to a nilpotent matrix also be nilpotent? If so, with the same index?
- Answer
Yes. Generalize the last sentence in Example 2.9. As to the index, that same last sentence shows that the index of the new matrix is less than or equal to the index of , and reversing the roles of the two matrices gives inequality in the other direction.
Another answer to this question is to show that a matrix is nilpotent if and only if any associated map is nilpotent, and with the same index. Then, because similar matrices represent the same map, the conclusion follows. This is Exercise 14 below.
- This exercise is recommended for all readers.
- Problem 9
Show that the only eigenvalue of a nilpotent matrix is zero.
- Answer
Observe that a canonical form nilpotent matrix has only zero eigenvalues; e.g., the determinant of this lower-triangular matrix
is , the only root of which is zero. But similar matrices have the same eigenvalues and every nilpotent matrix is similar to one in canonical form.
Another way to see this is to observe that a nilpotent matrix sends all vectors to zero after some number of iterations, but that conflicts with an action on an eigenspace unless is zero.
- Problem 10
Is there a nilpotent transformation of index three on a two-dimensional space?
- Answer
No, by Lemma 1.3 for a map on a two-dimensional space, the nullity has grown as large as possible by the second iteration.
- Problem 11
In the proof of Theorem 2.13, why isn't the proof's base case that the index of nilpotency is zero?
- Answer
The index of nilpotency of a transformation can be zero only when the vector starting the string must be , that is, only when is a trivial space.
- This exercise is recommended for all readers.
- Problem 12
Let be a linear transformation and suppose is such that but . Consider the -string .
- Prove that is a transformation on the span of the set of vectors in the string, that is, prove that restricted to the span has a range that is a subset of the span. We say that the span is a -invariant subspace.
- Prove that the restriction is nilpotent.
- Prove that the -string is linearly independent and so is a basis for its span.
- Represent the restriction map with respect to the -string basis.
- Answer
- Any member of the span can be written as a linear combination . But then, by the linearity of the map, is also in the span.
- The operation in the prior item, when iterated times, will result in a linear combination of zeros.
- If then the set is empty and so is linearly independent by definition. Otherwise write and apply to both sides. The right side gives while the left side gives ; conclude that . Continue in this way by applying to both sides, etc.
- Of course, acts on the span by acting on
this basis as a single, -long, -string.
- Problem 13
Finish the proof of Theorem 2.13.
- Answer
We must check that is linearly independent where is a -string basis for , where is a basis for , and where . Write
and apply .
Conclude that the coefficients are all zero as is a basis. Substitute back into the first displayed equation to conclude that the remaining coefficients are zero also.
- Problem 14
Show that the terms "nilpotent transformation" and "nilpotent matrix", as given in Definition 2.6, fit with each other: a map is nilpotent if and only if it is represented by a nilpotent matrix. (Is it that a transformation is nilpotent if an only if there is a basis such that the map's representation with respect to that basis is a nilpotent matrix, or that any representation is a nilpotent matrix?)
- Answer
For any basis , a transformation is nilpotent if and only if is a nilpotent matrix. This is because only the zero matrix represents the zero map and so is the zero map if and only if is the zero matrix.
- Problem 15
Let be nilpotent of index four. How big can the rangespace of be?
- Answer
It can be of any size greater than or equal to one. To have a transformation that is nilpotent of index four, whose cube has rangespace of dimension , take a vector space, a basis for that space, and a transformation that acts on that basis in this way.
--possibly other, shorter, strings--
So the dimension of the rangespace of can be as large as desired. The smallest that it can be is one— there must be at least one string or else the map's index of nilpotency would not be four.
- Problem 16
Recall that similar matrices have the same eigenvalues. Show that the converse does not hold.
- Answer
These two have only zero for eigenvalues
but are not similar (they have different canonical representatives, namely, themselves).
- Problem 17
Prove a nilpotent matrix is similar to one that is all zeros except for blocks of super-diagonal ones.
- Answer
A simple reordering of the string basis will do. For instance, a map that is assoicated with this string basis
is represented with respect to by this matrix
but is represented with respect to in this way.
- This exercise is recommended for all readers.
- Problem 18
Prove that if a transformation has the same rangespace as nullspace. then the dimension of its domain is even.
- Answer
Let be the transformation. If then the equation shows that is even.
- Problem 19
Prove that if two nilpotent matrices commute then their product and sum are also nilpotent.
- Answer
For the matrices to be nilpotent they must be square. For them to commute they must be the same size. Thus their product and sum are defined.
Call the matrices and . To see that is nilpotent, multiply , and , etc., and, as is nilpotent, that product is eventually zero.
The sum is similar; use the Binomial Theorem.
- Problem 20
Consider the transformation of given by where is an matrix. Prove that if is nilpotent then so is .
- Answer
Some experimentation gives the idea for the proof. Expansion of the second power
the third power
and the fourth power
suggest that the expansions follow the Binomial Theorem. Verifying this by induction on the power of is routine. This answers the question because, where the index of nilpotency of is , in the expansion of
for any at least one of the and has a power higher than , and so the term gives the zero matrix.
- Problem 21
Show that if is nilpotent then is invertible. Is that "only if" also?
- Answer
Use the geometric series: . If is the zero matrix then we have a right inverse for . It is also a left inverse.
This statement is not "only if" since
is invertible.